# How to graph a parabola (y-2)^2 =8x ?

May 13, 2015

First, let's divide both sides by 8 to get:

$x = {\left(y - 2\right)}^{2} / 8$

${\left(y - 2\right)}^{2} \ge 0$ for all values of $y$, but will equal $0$ when $y = 2$.

So the parabola has its vertex to the left on the $y$ axis at $\left(0 , 2\right)$ and its axis is a horizontal line parallel to the $x$ axis with formula $y = 2$. It is symmetric about this axis. The means that for every point $\left(x , y\right)$ that lies on the parabola, so does the point $\left(x , \left(4 - y\right)\right)$.

The lower 'arm' of the parabola will intersect the $x$ axis when $y = 0$. So substitute $y = 0$ in the equation of the parabola to derive $x$ ...

$x = {\left(y - 2\right)}^{2} / 8 = {\left(0 - 2\right)}^{2} / 8 = {\left(- 2\right)}^{2} / 8 = \frac{4}{8} = \frac{1}{2}$

In other words, the parabola intersects the $x$ axis at $\left(\frac{1}{2} , 0\right)$.

If you want to know any more points the parabola passes through, just choose a value of $y$ and substitute it into the formula for $x$ to get the corresponding value of $x$.